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Section 5.3 – The Definite Integral
As the number of rectangles increased, the approximation of the area under the curve approaches a value.
Copyright 2010 Pearson Education, Inc.
Section 5.3 – The Definite IntegralDefinition
Note: The function f(x) must be continuous on the interval [a, b].
Section 5.3 – The Definite IntegralParts of the Definite Integral
Copyright 2010 Pearson Education, Inc.
Section 5.3 – The Definite IntegralProperties of the Definite Integral
Copyright 2010 Pearson Education, Inc.
Section 5.3 – The Definite IntegralGeometric Interpretations of the Properties of the Definite Integral
Copyright 2010 Pearson Education, Inc.
Section 5.3 – The Definite IntegralUsing the Properties of the Definite Integral
∫1
3
𝑓 (𝑥 )𝑑𝑥=6∫3
7
𝑓 (𝑥 )𝑑𝑥=9∫1
3
𝑔 (𝑥 )𝑑𝑥=−4
∫1
3
3 𝑓 (𝑥 )𝑑𝑥=¿3∫1
3
𝑓 (𝑥 ) 𝑑𝑥=¿3 (6 )=18
∫1
3
(2 𝑓 (𝑥 )−4𝑔 (𝑥 ) )𝑑𝑥=¿2∫1
3
𝑓 (𝑥 )𝑑𝑥−4∫1
3
𝑔 (𝑥 )𝑑𝑥=¿2 (6 )−4 (−4 )=28
∫1
7
𝑓 (𝑥 )𝑑𝑥=¿∫1
3
𝑓 (𝑥 )𝑑𝑥+∫3
7
𝑓 (𝑥 )𝑑𝑥=¿6+9=15
∫3
1
𝑓 (𝑥 )𝑑𝑥=¿−∫1
3
𝑓 (𝑥 )𝑑𝑥=¿−6
Given:
Section 5.3 – The Definite IntegralRules of the Definite Integral
∫𝑎
𝑏
𝑐𝑑𝑥=𝑐 (𝑏−𝑎) ∫𝑎
𝑏
𝑥 𝑑𝑥=𝑏2
2−𝑎2
2∫𝑎
𝑏
𝑥2𝑑𝑥=𝑏3
3−𝑎3
3
Examples
∫2
6
4𝑑𝑥=¿¿ 4 (6−2 )=¿16 ∫4
8
𝑥 𝑑𝑥=¿¿ 82
2−
42
2=¿32−8=¿24
∫3
5
𝑥2𝑑𝑥=¿53
3−
33
3=¿
1253−
273
=¿9 83
=32.67
∫3
4
𝑥2+3 𝑥−2𝑑𝑥=¿∫3
4
𝑥2𝑑𝑥+3∫3
4
𝑥 𝑑𝑥−∫3
4
2𝑑𝑥=¿43
3−
33
3+¿3 ( 42
2−
32
2 )−2 (4−3 )=¿
643−
273
+¿3 (162−
92 )−2 (1 )=¿20.83
Section 5.4 – The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus
∫𝑎
𝑏
𝑓 (𝑥 )𝑑𝑥=𝐹 (𝑏)−𝐹 (𝑎) .
Examples
∫1
5
5 𝑥 𝑑𝑥=¿5𝑥2
2=¿
5
1
5(5)2
2−
5 (1 )2
2=¿
1252−
52=¿120
2=60
∫𝜋6
2𝜋3
𝑠𝑖𝑛𝑥 𝑑𝑥=¿−𝑐𝑜𝑠𝑥=¿5
1
−𝑐𝑜𝑠 ( 2𝜋3 )−(−𝑐𝑜𝑠( 𝜋6 ))=¿−(− 1
2 )−(− √32 )=¿0 .866
Section 5.4 – The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus
∫𝑎
𝑏
𝑓 (𝑥 )𝑑𝑥=𝐹 (𝑏)−𝐹 (𝑎) .
∫3
4
𝑥2+3 𝑥−2𝑑𝑥=¿
Examples
𝑥3
3+ 3𝑥2
2−2𝑥=¿
4
3
43
3+
3 (4)2
2−2(4 )−( 33
3+
3(3)2
2−2(3) )
37.33−16.5=¿20.83
∫1
321
𝑥65
𝑑𝑥=¿ 𝑥−15
−15
=¿
32
1
−5
𝑥15
=¿−52−(−5
1 )=¿2.5∫1
32
𝑥−65 𝑑𝑥=¿
32
1
Section 5.4 – The Fundamental Theorem of CalculusDifferentiating a Definite Integral
𝐼𝑓 𝑓 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑜𝑛 [𝑎 ,𝑏 ]𝑎𝑛𝑑 𝑥𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑜𝑛 [𝑎 ,𝑏 ] , h𝑡 𝑒𝑛
𝐹 ′ (𝑥 )= 𝑑𝑑𝑥∫𝑎
𝑥
𝑓 (𝑡 ) 𝑑𝑡= 𝑓 (𝑥)
𝑑𝑑𝑥∫1
𝑥
𝑡 2𝑑𝑡=¿¿ 𝑥2𝑑𝑑𝑥 ( 𝑡
3
3 )=¿𝑥
1
𝑑𝑑𝑥 (𝑥
3
3−
13
3 )=¿𝑑𝑑𝑥 (𝑥
3
3−
13 )=¿
𝑑𝑑𝑥∫1
4𝑥
𝑡 2𝑑𝑡=¿¿ 64 𝑥2𝑑𝑑𝑥 ( 𝑡
3
3 )=¿4 𝑥
1
𝑑𝑑𝑥 ((4 𝑥)3
3−
13
3 )=¿𝑑𝑑𝑥 ( 64 𝑥3
3−
13 )=¿
𝑑𝑑𝑥∫1
𝑥2
𝑡 2𝑑𝑡=¿¿ 6 𝑥5
3=¿𝑑
𝑑𝑥 ( 𝑡3
3 )=¿𝑥2
1
𝑑𝑑𝑥 ((𝑥2)3
3−
13
3 )=¿𝑑𝑑𝑥 (𝑥
6
3−
13 )=¿ 2 𝑥5
Section 5.4 – The Fundamental Theorem of CalculusDifferentiating a Definite Integral
𝐼𝑓 𝑓 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑜𝑛 [𝑎 ,𝑏 ]𝑎𝑛𝑑 𝑥𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑜𝑛 [𝑎 ,𝑏 ] , h𝑡 𝑒𝑛
𝐹 ′ (𝑥 )= 𝑑𝑑𝑥∫𝑎
𝑥
𝑓 (𝑡 ) 𝑑𝑡= 𝑓 (𝑥)
𝑑𝑑𝑥∫1
𝑥
𝑡 2𝑑𝑡=¿¿(𝑥 )2 (1 )=¿
𝑑𝑑𝑥∫1
4𝑥
𝑡 2𝑑𝑡=¿¿ 64 𝑥2
𝑑𝑑𝑥∫1
𝑥2
𝑡 2𝑑𝑡=¿¿(𝑥2 )2 (2𝑥 )=¿
𝑥2
( 4 𝑥 )2 ( 4 )=¿
2 𝑥5
𝑑𝑑𝑥∫1
𝑥2
sin 𝑡 𝑑𝑡=¿¿ (2 𝑥 )𝑠𝑖𝑛 (𝑥2 )
𝑑𝑑𝑥 ∫
1
𝑥2+𝑥
tan 𝑡 𝑑𝑡=¿¿ (2 𝑥+1 ) 𝑡𝑎𝑛 (𝑥2+2 )
𝑑𝑑𝑥 ∫
𝑡𝑎𝑛𝑥
𝑠𝑖𝑛𝑥
𝑡 2𝑑𝑡=¿¿(𝑠𝑖𝑛¿¿2𝑥) (𝑐𝑜𝑠𝑥 )¿−(𝑡𝑎𝑛¿¿ 2𝑥) (𝑠𝑒𝑐2𝑥 )¿
Section 5.4 – The Fundamental Theorem of CalculusMean Value Theorem for Integrals
Copyright 2010 Pearson Education, Inc.
Section 5.3 – The Definite IntegralMean Value for Definite Integral
Find the mean (average) value of the function over the interval
𝐴𝑉= 13−1
∫1
3
(𝑥2+2 )𝑑𝑥
𝐴𝑉=12 ( 𝑥
3
3+2𝑥)
𝐴𝑉=12 [( 27
3+6)−( 1
3+2)]
𝐴𝑉=12 [15−
73 ]
𝐴𝑉=193
=6.33
3
1
Section 5.3 – The Definite IntegralDifference between the Value of a Definite Integral and Total Area
Find the mean (average) value of the function over the interval
Copyright 2010 Pearson Education, Inc.
Value of the Definite Integral
∫0
2 𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥=¿−𝑐𝑜𝑠𝑥=¿2𝜋
0
−𝑐𝑜𝑠 (2𝜋 )− (−𝑐𝑜𝑠 (0 ) )=¿
− (1 )− (−1 )=¿0
Total Area
∫0
𝜋
𝑠𝑖𝑛𝑥𝑑𝑥−∫𝜋
2 𝜋
𝑠𝑖𝑛𝑥𝑑𝑥=¿
−𝑐𝑜𝑠𝑥−𝜋
0
−𝑐𝑜𝑠 (𝜋 )− (−𝑐𝑜𝑠 (0 ) )−− (−1 )− (−1 )−
4
∫0
2 𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥=¿
[(−𝑐𝑜𝑠 (2𝜋 )− (−𝑐𝑜𝑠 (𝜋 ) ) ) ]
−𝑐𝑜𝑠𝑥2𝜋
𝜋
(− (1 )− (1 ) )=¿